Problems

# Find equation of motion using. Lagrangian given equation is $L'=\frac{m}{2}(a\dot{x}^2+2b\dot{x}{y}+c\dot{y}^2)-\frac{k}{2}(ax^2+2bxy+cy^2)$

+1
−0

$$L'=\frac{m}{2}(a\dot{x}^2+2b\dot{x}{y}+c\dot{y}^2)-\frac{k}{2}(ax^2+2bxy+cy^2)$$
where a, b, and c are arbitrary constants but subject to the condition that $b^2 − ac \ne 0$. What are the equations of motion? Examine particularly the two cases $a = 0 = c$ and $b = 0$, $c = −a$. What is the physical system described by the above Lagrangian? Show that the usual Lagrangian for this system as defined by Eq. (1.56) is related to $L'$ by a point transformation (cf. Derivation 10). What is the significance of the condition on the value of $b^2 − ac$?

$L=T-U \ \text{Eq.(1.56)}$

It's two dimensional. If there was only $x$ than, it was easier to differentiate. But, I was thinking what I should do now. Then, all the things I had done that is differentiating respect to $x$ and $y$ both.

$$\frac{d^2 L}{dx dy}=-kax-bkx-bky-cky$$ $$\frac{d^2 L}{d\dot{x}d\dot{y}}=am\dot{x}+bm\dot{y}+bm\dot{x}+cm\dot{y}$$ $$\frac{d}{dt}(\frac{d^2 L}{d\dot{x}d\dot{y}})=am\ddot{x}+bm\ddot{y}+bm\ddot{x}+cm\ddot{y}$$ $$\frac{d}{dt}(\frac{d^2 L}{d \dot{x} d\dot{y}})-\frac{d^2}{dxdy}=0$$ $$am\ddot{x}+bm\ddot{y}+bm\ddot{x}+cm\ddot{y}=-(kax+bkx+bky+cky)$$ How can I find equation of motion for that? (What are the equations of motion?) Should I find a equation for $x$ at first then another equation for $y$?

Why does this post require moderator attention?
Why should this post be closed?

+2
−0

Your mistake is that you did a second derivative of $L$, taking the derivative according to both degrees of freedom together. Instead you need to make a separate equation for each degree of freedom.

Since you have two degrees of freedom ($x,y$), you get two equations:

\begin{align} \frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial\dot x} - \frac{\partial L}{\partial x} &= 0\\ \frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial\dot y} - \frac{\partial L}{\partial y} &= 0 \end{align}

The two equations you get this way are the equations of motion.

Another mistake is that you are writing total derivatives where you would need to write partial ones; the only total derivative is the time derivative. Since you are using them as if they were partial derivatives, this is inconsequential in your calculation, though.

Why does this post require moderator attention?